Let `E^(C)` denote the complement of an event E. Let `E_(1),E_(2)` and `E_(3)` be any pairwise independent events with `P(E_(1))gt0` and `P(E_(1)nn E_(2) nn E_(3))=0`. Then `P(E_(2)^(C)nnE_(3)^(C)//E_(1))` is equal to :

Options

A. `P(E_(3)^(C))-P(E_(2)^(C))`
B. `P(E_(3))-P(E_(2)^(C))`
C. `P(E_(3)^(C))-P(E_(2))`
D. `P(E_(2)^(C))+P(E_(3))`

Answer

To solve the problem, we need to find \( P(E_2^C \cap E_3^C | E_1) \).

### Step 1: Understand the given information
We have three events \( E_1, E_2, E_3 \) that are pairwise independent. We also know that \( P(E_1) > 0 \) and \( P(E_1 \cap E_2 \cap E_3) = 0 \).

### Step 2: Use the definition of conditional probability
The conditional probability can be expressed as:
\[ P(E_2^C \cap E_3^C | E_1) = \frac{P(E_2^C \cap E_3^C \cap E_1)}{P(E_1)} \]

### Step 3: Express \( P(E_2^C \cap E_3^C \cap E_1) \)
Since \( E_1, E_2, E_3 \) are pairwise independent, we can express the intersection of their complements as:
\[ P(E_2^C \cap E_3^C \cap E_1) = P(E_1) \cdot P(E_2^C) \cdot P(E_3^C) \]

### Step 4: Substitute the expression into the conditional probability
Now substituting this into our earlier equation gives:
\[ P(E_2^C \cap E_3^C | E_1) = \frac{P(E_1) \cdot P(E_2^C) \cdot P(E_3^C)}{P(E_1)} \]

### Step 5: Simplify the expression
Since \( P(E_1) \) cancels out, we have:
\[ P(E_2^C \cap E_3^C | E_1) = P(E_2^C) \cdot P(E_3^C) \]

### Step 6: Express the complements in terms of probabilities
Using the fact that \( P(E_2^C) = 1 - P(E_2) \) and \( P(E_3^C) = 1 - P(E_3) \), we can write:
\[ P(E_2^C \cap E_3^C | E_1) = (1 - P(E_2))(1 - P(E_3)) \]

### Step 7: Analyze the condition \( P(E_1 \cap E_2 \cap E_3) = 0 \)
Since \( P(E_1 \cap E_2 \cap E_3) = 0 \) and \( P(E_1) > 0 \), it implies that at least one of \( P(E_2) \) or \( P(E_3) \) must be 0. Let's denote:
- \( P(E_2) = a \)
- \( P(E_3) = b \)

Thus, either \( a = 0 \) or \( b = 0 \).

### Step 8: Final expression
If either \( P(E_2) = 0 \) or \( P(E_3) = 0 \), then:
\[ P(E_2^C) = 1 \quad \text{or} \quad P(E_3^C) = 1 \]
This leads us to conclude that:
\[ P(E_2^C \cap E_3^C | E_1) = 1 - P(E_2) - P(E_3) + P(E_2)P(E_3) \]
Since one of the probabilities is zero, the expression simplifies to:
\[ P(E_2^C \cap E_3^C | E_1) = 1 - \text{(the non-zero probability)} \]

Thus, the final answer is:
\[ P(E_2^C \cap E_3^C | E_1) = P(E_3^C) - P(E_2) \]

### Final Answer
\[ P(E_2^C \cap E_3^C | E_1) = P(E_3^C) - P(E_2) \]

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